2

We always hear of 4 Forces or interactions, i.e. the Strong Force, Electromagnetic Force, Weak Force and Gravity. Each force has its mediating Bosons, i.e. gluons for Strong Force, photons for Electromagnetic Force, and theorized gravitons for Gravity. However, only the so called Weak Force has 2 types of bosons, i.e. the W (+ or -) bosons and the Z boson.

Moreover, the type of interactions mediated by the W and Z bosons are quite different, some of which differences are:

  1. W bosons mediate flavour changing decays, from one type of fundamental particle to another. In fact this is the only interaction that can modify a fundamental particle (not sure the particles are so fundamental, but that's another question). However, Z bosons mediate elastic scattering between neutrinos and other fundamental particles such as electrons (not sure if they also mediate elastic scattering between other fundamental particles, e.g. an electron and another electron or an electron and a quark, but there would be other forces between such particles, which would perhaps drown out Z boson mediated elastic scattering, if any).
  1. The W bosons and Z boson interactions have different charges, the Weak Isospin for W bosons and Weak Charge for Z bosons. Also see https://physics.stackexchange.com/a/262300/359335

  2. W bosons interactions violate parity conservation, while Z boson interactions don't violate parity conservation.

  3. W bosons interact only with left handed leptons or quarks or right handed anti-particles, while Z bosons can interact with right handed leptons and quarks, and with left handed anti-particles, since right handed particles also carry Weak Charge (see Wikipedia link above).

Even otherwise each force carrying boson has its own unique interaction, with the newly discovered Higgs Boson also being responsible for the mass generating interactions (which interactions also mix together the left and right handed particles). Therefore, why do interactions mediated by the W bosons and Z boson share the same name?

I have read some articles on electroweak symmetry breaking and realise that the photon, W bosons and the Z boson are mixtures of the original four massless bosons prior to symmetry breaking, and that these forces unite at higher energies. I also realise that the Electric Charge, Weak Charge and Weak Isospin are some formulaic mixtures of each other and of the Weak Hypercharge.

However, that really doesn't explain why the W bosons and Z boson interaction are called out together, when in the real world they are quite different, at least as different as the photon and Z boson interactions (Why are the Z boson and photon different?).

The only reasons I was able to come up with, for the same name, is history and convenience? The Weak interaction as it was originally discovered and studied was apparently through Beta decays. Therefore, it was initially thought to be only a single type of an interaction i.e. the interaction, which is now known to be mediated by W bosons (W bosons were unknown at that time). Subsequent formulation of the electroweak theory by Sheldon Glashow, Abdus Salam and Steven Weinberg, led to prediction of the W bosons and Z boson, and search for the Z boson mediated elastic scattering and neutral currents, which were subsequently discovered experimentally (in a huge success for the Standard Model). However, despite the two Weak interactions being different in the real world, no new name was thought of for the Z boson interaction and it was also clubbed with the Weak Force, since it was also a short distance interaction mediated by a heavy boson, just like the already known Weak interaction.

Sorry for a long question, but I wanted to make clear that I am not looking for mathematics on how electroweak interactions are related, which fact I am implicitly accepting. I am only asking that if the Electromagnetic and W bosons interactions are named separately, why not the W bosons and Z boson interactions?

Suneet
  • 21
  • 1
    Because we have a theory that encompasses both and the same coupling constant. See https://en.wikipedia.org/wiki/Electroweak_interaction – ohwilleke Mar 10 '23 at 20:32
  • 1
    Obligatory xkcd - if you think the "force" in "four fundamental forces" relates directly in any way to a classical/intuitive idea of force, you're gonna have a bad time. – ACuriousMind Mar 10 '23 at 22:41
  • 3
    Your 3. is *completely wrong* . Z interactions do not preserve parity--they just don't violate it maximally. Try a book or a review or WP. If you can err on this, you have failed to see the subtle method in the madness of electroweak unification. I doubt a short question could possibly deconstruct and reconstruct your flawed thinking. – Cosmas Zachos Mar 10 '23 at 23:23
  • @ACuriousMind Indeed. But turning physics into a branch of mathematics disconnected from reality is not a good thing. – John Doty Mar 11 '23 at 13:55

3 Answers3

5

First, interactions of the Z absolutely violate parity. Note how the Z boson couples to fermions:

$-i \frac{g}{\cos{\theta_{W}}} \bar{\psi}_{f}\gamma^{\mu}(v_{f} - a_{f}\gamma^{5})\psi_{f}$

where $v_{f} = T_{3,f} - 2Q_{f}\sin^{2}\theta_{W}$, and $a_{f} = T_{3,f}$. Here $Q_{f}$ is the charge of the fermion and $T_{3}$ is +1/2 or -1/2. The asymmetry of the Z decay at the Z pole is a huge input to electroweak precision analysis.

Secondly the Z and W bosons are intimately related because of what you alluded, that they come from the spontaneous symmetry breaking of SU(2)xU(1). If you don't want mathematical explanations I'm not sure what else can please you, but note the fact that the photon is massless and mediates a long range force because of this. This W and Z are both massive (with masses related by symmetry breaking) and thus both mediate short range forces with the same strength.

CStarAlgebra
  • 2,262
  • 1
  • 14
  • 25
  • Thanks. What I understand from above is that W and Z bosons are both massive with similar masses (~90 GeV and ~80 GeV respectively), and therefore interactions mediated by the 2 bosons are of similar strength and range. Hence, they are clubbed together as the Weak Force. Also, I understand that coupling strength of Z boson to left and right handed fundamental particles is different, thereby violating parity (e.g. in Z boson decays). I perhaps framed my question incorrectly, as I meant there is no change of parity between initial and final products in an elastic scattering mediated by Z bosons. – Suneet Mar 13 '23 at 10:03
3

$W$ bosons are a combination of 2 of the SU(2) generators ($W_1$,$W_2$), while Z and the photon a mix between the 3rd generator ($W_3$) of SU(2) and the one of U(1), $B$. Their common origin results in the strength of the interactions with Z and W bosons to be also tied to one another through the weak angle or Weinberg angle, $\theta_W$. As you also acknowledge, weak hypercharge (coupling to $B$), (third component of) weak isospin and electric charge are connected. While charged currents don't couple to right-handed particles at all, neutral currents do, but still in a different way to left-handed ones (as opposed to the photon, which does not distinguish between left and right chirality).

To answer your final question - "I am only asking that if the Electromagnetic and W bosons interactions are named separately, why not the W bosons and Z boson interactions?" - what separates the photon from the rest is that a combination of the generators in the GWS model is unbroken (the one which corresponds to the photon, coupling to electric charge, $SU(2) \times U(1)\rightarrow U(1)_{em}$), and hence the photon does not acquire a mass term by interactions with the Higgs field. You can also argue that there is also a historical reason - QED was developed before the formulation of EW theory. Separation also depends on which energy scale you are working in - at a point, there is no electromagnetic interaction anymore.

Vangi
  • 592
  • Thanks for the answer. An additional reason to club the two interactions together (in addition to what @CStarAlgebra clarified above) is because Z boson and W boson interactions decouple from the Electromagnetic interaction at the same energy scale? – Suneet Mar 13 '23 at 10:08
  • 1
  • I am not sure I understand your comment. The $Z$, $W$ and the photon arise through the same mechanism (so "at the same energy scale" as you say). What I think @CStarAlgebra is saying is that the $W$ and the $Z$ are not considered as separate forces because they are a result of the same mechanism (the breaking of the $SU(2)\times SU(1)$ symmetry), and while the photon remains massless, those two do acquire mass. This is more or less what I have in the answer as well (although I did not explicitly say $W$ and $Z$ acquire mass).
  • – Vangi Mar 14 '23 at 18:17
  • 1
  • What @CStarAlgebra points out in addition is that due to their similar masses (linked to the symmetry breaking energy scale and related to one another through the angle I mentioned), the interactions they mediate are short ranged compared to the interactions of the massless photon.
  • – Vangi Mar 14 '23 at 18:21